Anmeldelse af “The Four Sisters Overlooking the Sea” (gratis, hvis linket stadig virker), af #NaomiKritzer. (Link nedenfor.) Langnovelle. 2024. Hugo-finalist.

Skitse: Hun er vel sådan set mest en husmor, der fx kører datteren til og fra teatret, når der er en musical på tapetet. Hun er gift med en karriere-akademiker, der typisk lader hende redigere sine artikler, før de bliver offentliggjort. Omend ordet “redigere” her omfatter at skrive det meste af teksten … Så bor de også i nærheden af en klippeformation, man heromkring kalder de 4 søstre.

Er det science fiction? Nej da. Fantasy.

Temaer: Hun havde også selv en akademisk karriere, hvor hun studerede sæler. Og det er altså underligt, men selvom hun er flyttet meget langt, så er der stadig de samme sæler ved vandet?

Hvis det ikke allerede fremgår tydeligt: ægtemand = r*vhul.

Er det godt? Jeg føler absolut med hende. Yeah, kvinder der gør fremskridt i verden. 👽👽👽

***

“The Four Sisters Overlooking the Sea”

Anmeldelse af “The Brotherhood of Montague St. Video” (gratis), af Thomas Ha. (Link nedenfor.) Langnovelle. 2024. Hugo-finalist.

Skitse: I den her verden er det underligt, når en bog ikke kan zoome eller blive lysere. Det er også usædvanligt, at man endnu kan få konverteret en gammel video, uden at få den redigeret.

Er det science fiction? Ja.

Temaer: Selvfølgelig er der fordele ved de elektroniske bøger. Gammeldags ord kan blive opdateret, og man kan justere lysstyrken. Men (som sædvanlig?) har det taget overhånd. En trist slutning bliver “opdateret”.

Her er et anstrøg af Fahrenheit 451.

Det næste er så, når personlige minder bliver ændret. Fordi hvem gider have en familievideo, der er trist?

For den sags skyld bliver hele livet set gennem briller, der lige hæver det generelle niveau på alting. Kærestens ansigt bliver lidt kønnere.

Livet bliver mindre kompliceret. Mere kedeligt.

Citat: “What is it these people want? What is it these people need? Are they striving toward one, or the other, with what they do each day?”

Er det godt? Den nye verden, hvor bøger ikke bliver brændt, bare justeret. Oh. Ikke helt kreativt nok til at imponere mig. 👽👽☠️

***

“The Brotherhood of Montague St. Video” by Thomas Ha

Anmeldelse af “Signs of Life” (gratis), af Sarah Pinsker. (Link nedenfor.) Langnovelle. 2024. Hugo-finalist.

Skitse: 45 år er længe nok. Nu er det på tide at opsøge sin søster og sige undskyld. Hun er ensom, hendes mand og 3 børn er døde, og kun Shane er tilbage.

Er det science fiction? Fantasy.

Temaer: Selvfølgelig handler det om Det Store Familieskænderi, at blive gammel, endelig få sagt undskyld osv.

Men der er noget mere. En dybere årsag til skænderiet. At søsteren kan noget, æhm, magisk. Og at hun har brug for hjælp, sådan som et ældre, ensomt familiemedlem kan have.

Citat: “I forgave you years and years and years ago. I tried to tell you at Father’s funeral, but I never managed to get near you. That’s when I understood that you still needed to forgive yourself, too. I think that’s what’s taken a little longer.”

Er det godt? Velskrevet. Det er ikke underligt med den her forfatter. Ca. halvvejs er der en afsløring, der nærmest gav mig et stød. Slet ikke så dårligt. 👽👽👽

***

“Signs of Life”, Sarah Pinsker

Anmeldelse af “Loneliness Universe” (gratis), af Eugenia Triantafyllou. (Link nedenfor.) Langnovelle. 2024. Hugo-finalist.

Skitse: Nefeli beslutter at kontakte en gammel ven, hun ikke rigtig har forbindelse med mere. De aftaler at mødes ved et busstoppested, men et eller andet går galt.

Er det science fiction? Det er jo lidt en fortolkning. Men svaret kan godt være ja.

Temaer: Nefeli oplever i stigende grad, at hun ikke kan være i stue sammen med sine venner og bekendte. Helt bogstaveligt. Hun kan sende en tekst-besked til sin bror, og hun kan spise de chips, han lige har lagt på bordet, men derudover er det, som om han lige har forladt lokalet. For ham er det, som om hun lige er gået.

Fænomenet spreder sig, og videnskabsfolk prøver at finde årsagen. Nefeli har også et bud.

Det er her, det muligvis mere bliver symbolsk det hele. At dagsordenen er at skælde ud på folk, der fysisk er sammen, men faktisk har næsen i mobilen, hver for sig.

Er det godt? Jeg var spændt på slutningen, på opklaringen. Som altid er det lidt irriterende, når fortolkningen ikke er entydig. 👽👽☠️

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“Loneliness Universe”, Eugenia Triantafyllou

Anmeldelse af “Lake of Souls”, af Ann Leckie. Langnovelle. 2024. Hugo-finalist.

Skitse: Via at skifte ham har lille Spawn allerede været gennem faserne æg og larve og kan se frem til at blive rigtig voksen med et navn næste gang. Han er optaget af spørgsmålet om væsner uden sjæl.

Er det science fiction? Ja.

Temaer: Der er også et menneske, en antropolog, der studerer livet på den her fremmede planet. Han har ellers været frosset ned i lang tid, men nu er han her og prøver at undgå alt for tæt kontakt med de lyserøde kravlefiduser, mens han leder efter ekspeditionens ansible.

Spawn foretager en lang rejse. Da han støder på antropologen, er det afgørende, at de hver især kan se, at den anden har en taske. Taske = intelligens.

Er det godt? Det kører egentlig meget godt. Var det nødvendigt at have et menneske med? Jeg var spændt på slutningen. 👽👽👽

Anmeldelse af “By Salt, by Sea, by Light of Stars” (gratis), af Premee Mohamed. Langnovelle. 2024. Hugo-finalist.

Skitse: Firion kan stor magi! Men på det sidste har hun været lidt distraheret, fordi hendes magiske batterier er flade. Hun har ikke fået tjekket sin inbox, så det er en overraskelse, da der står en elev ved døren.

Er det science fiction? Nej da. Fantasy.

Temaer: Firion skal have det hele til at gå op i en højere enhed. Få sin magi tilbage, forhåbentlig. Uddanne eleven. Beskytte den nærliggende landsby mod en superdrage. Der bliver i hvert fald en historie ud af det.

(Det er sært, lige at have læst noget indviklet, hvor man skal lede efter meningen, og så læse noget, der er ret lige ud ad landevejen. Fordi det er jo bare forskelligt, ikke dårligt. Ikke?)

Er det godt? Det virker. Men nok ikke stor kunst. 👽👽☠️

This week the #puzzle is: Happy 100th Fiddler! #counting #permutations #triangles #parallelograms #dozo #flag

Dozo is a strategy game with a rather distinctive board:

The board features 28 holes in which players place markers, with the goal of making an equilateral triangle of any size with one color.

How many distinct equilateral triangles can you find whose vertices are the centers of holes on the board? (If two triangles are congruent but have different vertices, they should still be counted as distinct.)

And for extra credit:

Happy Fourth of July! In celebration of America’s birthday, let’s count more shapes—not in a board game, but in the American flag:

In particular, consider the centers of the 50 stars depicted on the flag. How many distinct parallelograms can you find whose vertices are all centers of stars? (If two parallelograms are congruent but have different vertices, they should still be counted as distinct.)

Highlight to reveal (possibly incorrect) solution:

Program Handcount 1 and 2

I tried counting triangles both by hand and with a program. The handcount illustrated is actually a little wrong, because in a certain case (6 + 6) I forgot to count the triangles, where the tip pointed up. The program goes through all the options for the 1st and 2nd vertex, constructing the 3rd and testing whether this hypothetical 3rd is on the board. This was a little tricky, because my model of the board didn’t preserve angles and distances from the board. But I think I got there in the end. Result: 126 triangles.

And for extra credit:

Program

This time I went straight to the program. It goes through all the options for 3 vertices, constructing the 4th. There’s actually a charming set of formulas for constructing the 3 options for the 4th vertex. This time it didn’t matter that angles and distances weren’t preserved. Only tricky part was to detect, when the 1st, 2nd and 3rd vertex were on a line. Result: 5918 parallelograms.

***

For the 3rd time I’ve come back to weaving the web. This time because a friend kindly helped me turn a 3d object file into a physical object:

This week the #puzzle is: Can You Crack the Roman Code? #combinations #recursion (Link at the bottom.)

And for extra credit:

Highlight to reveal (possibly incorrect) solution:

Tribonacci numbers Program

What do we know?

1. If the scroll had 1 * I, there would be 1 solution.
2. If it had 2, there would be 2 solutions, (I,I) and (II).
3. If it had 3, there would be 4 solutions, (I,I,I), (II,I), (I,II) and (III).
4. If it had 4, there would be 3 cases:
   • All the solutions for 3 * I with (I) added.
   • All the solutions for 2 * I with (II) added.
   • All the solutions for 1 * I with (III) added.
5. So this should be 4 + 2 + 1 = 7 solutions.
6. In general it’s the case, that the number of solutions f(n) = f(n-1) + f(n-2) + f(n-3).
7. These are actually the tribonacci numbers.
8. f(10) = a(12) = 274.

Just to make sure, I also wrote a recursive program to do the actual search through the tree of possibilities. Same answer.

And for extra credit:

Spreadsheet

Good thing I wrote that program! I can simply use it again. It says there are 4000 solutions.

Just to make sure, I also made a spreadsheet, counting in another way. Same result. I am happy.

***

Last week I actually didn’t do that well. I made a lot of assumptions, and some of them were wrong. In both puzzles my pass no. 2 didn’t correspond with the official solution. I’ve discovered, that for the fiddler this happened, because I made a mistake in my calculations. The area of my tilted 2nd pass is not 1.0472, but 0.7585. Oops.

***

Can You Crack the Roman Code?

This week the #puzzle is: How Greedily Can You Mow the Lawn? #geometry #area #volume #GreedyAlgorithm (Link at the bottom.)

You’re mowing a circular lawn with a radius of 1 unit. You can mow in straight strips that are 1 unit wide.

The fewest number of passes you would need to mow the entire lawn is two, as shown below. In one pass (shown in blue) you can mow half the circle, and in the second pass (shown in red) you can mow the other half of the circle.

However, instead of minimizing the number of passes, you greedily choose how to orient each pass so it cuts as much of the unmowed grass as possible. A pass doesn’t have to go through the center of the circle and can be in any direction, but must be straight and cannot bend.

With this “greedy” approach, how many passes will it take for you to mow the entire lawn?

And for extra credit:

Highlight to reveal (possibly incorrect) solution:

Desmos 1

Let us go through what choices we have.

• Obviously the 1st pass is simply symmetrical, the middle of the pass going through the center. If it didn’t go through the center, it would cover less area. In my model this pass is horizontal.
• The 2nd pass might simply be the same, just in the other direction? It would get a lot of the 2 remaining bits, leaving 4 bits behind. If this pass is indeed vertical, again, the middle of it goes through the center, to cover as much area as possible. This area would be 0.9132. [ETA: Oops, should be 2 * 0.9132. Wrong conclusions follow.]
• Then I would need 2 passes to cover the remaining 4 bits, either 2 horizontal (or vertical) passes, covering 2 bits every time, or 2 passes crisscrossing, one going up, the other going down, again covering 2 bits every time.
• That would be 4 passes in all.
• Can it be done in a better way?
• I need at least 3 passes, because I can’t cover the 2 remaining bits after the 1st pass in just 1 more pass.
• What if the 2nd pass simply covered 1 of the remaining bits? This area turns out to be 0.6142. So this wouldn’t happen, as it is less than 0.9132.
• There’s an interesting option. What if the 2nd pass was tilted just right, so that it wouldn’t create 4 remaining bits, but only 2? This covers an area of 1.0472. Yes! This is indeed possible. A 3rd pass, tilted the other way, would cover the rest.
• I don’t know whether this is exactly the optimal way to do things. But it proves, that 3 passes can do the job.

And for extra credit:

Desmos 2 Researchgate

There’s something interesting going on at Researchgate, suggesting that a non-greedy solution would need at least 20 circles / 10 cylinders. Because the angular radius of each circle is 30^o.

This one is just me guessing most of the way, illustrated in Desmod:

• 1st and 2nd pass are symmetrical, with the middle of the cylinder passing through the center. They are also perpendicular.
• I assume the same goes for the 3rd pass.
• The next 4 passes (red in my model) are closing some gaps on the surface of the sphere between the 3 existing passes. E.g. one of these cylinders has x=y=z in the middle.
• This leaves 8 bits. I would need 4 more passes to cover these, again using cylinders passing through the center.
• Using this method, with the whole surface covered and all cylinders passing through the center, all of the volume has been covered.
• This would suggest 11 passes.

***

How Greedily Can You Mow the Lawn?

This week the #puzzle is: Can You Race Against Zeno? #integration #summation #speed #distance #time (Link at the bottom.)

And for extra credit:

Highlight to reveal (possibly incorrect) solution:

Desmos 1 – just to be confusing, the definition of t_Zeno1 here is slightly different.

Let’s look at what we know or define:

• d: the full distance (5 km – all we metric centric people thank you)
• t_init: the time spent running with the initial speed all the way (24 m)
• v₀: the initial speed (d/t_init) (also, t_init = d/v₀)
• a: the increase of speed as a factor (1.1)
• t_Zeno1: the time spent running under the new rule

With these definitions in place, let’s go running.

• Run at v₀ to cover the distance d/2^1. (Half of the distance is run at the initial speed.)
• Then run at v₀ * a to cover the distance d/2². (Half of the remaining distance is run after 1 speed increase.)
• Then run at v₀ * a² to cover the distance d/2³.
• Then run at v₀ * a³ to cover the distance d/2⁴.
• In general run at v₀ * aⁱ to cover the distance d/2ⁱ⁺¹.
• In general spend t_i to cover distance d/2ⁱ⁺¹.
• t_i = d/2ⁱ⁺¹ / (v₀ * aⁱ)
• t_i = (d/v₀) / (2ⁱ⁺¹ * aⁱ)
• t_i = t_init / (2ⁱ⁺¹ * aⁱ)
• t_Zeno1 = Σ_i=0 t_i
• t_Zeno1 = Σ_i=0 t_init / (2ⁱ⁺¹ * aⁱ)
• t_Zeno1 = t_init * Σ_i=0 1 / (2ⁱ⁺¹ * aⁱ)
• To get a good approximation, I can let the upper bound be finite, but growing.
• According to Desmos, t_Zeno1 = 22 m.

But seeing I am so close, maybe I can do this in a better way.

• The general formula for v can be written as
• v(t) = a^{(⌊-log₂(1-t)⌋)} * v₀
• Some examples:
  • v(0) = a^{(⌊-log₂(1-0)⌋)} = a^{(⌊-log₂(1)⌋)} = a^(⌊0⌋) = 1 * v₀
  • v(0.4) = a^{(⌊-log₂(1-0.4)⌋)} = a^{(⌊-log₂(0.6)⌋)} = a⁰ = 1 * v₀
  • v(0.5) = a^{(⌊-log₂(1-0.5)⌋)} = a^{(⌊-log₂(0.5)⌋)} = a^{(⌊-(-1)⌋)} = a * v₀
• Instead of the sum I can use an integral.
• t_Zeno1 = t_init * ∫ ₀¹ 1 / (v(t)/v₀) dd
• Because, for a small distance d_j the contribution to the time is:
• t_j = d_j / v(t_j)
• t_Zeno1 = Σ_{all j} t_j
• t_Zeno1 = Σ_{all j} d_j / v(t_j)
• t_Zeno1 = Σ_{all j} d * d_j/d / v(t_j)
• t_Zeno1 = d * Σ_{all j} d_j/d / v(t_j)
• t_Zeno1 = d * ∫ ₀¹ 1 / v(t) dd
• t_Zeno1 = d * ∫ ₀¹ 1 / (a^{(⌊-log₂(1-t)⌋)} * v₀) dd
• t_Zeno1 = d/v₀ * ∫ ₀¹ 1 / a^{(⌊-log₂(1-t)⌋)} dd
• t_Zeno1 = t_init * ∫ ₀¹ 1 / a^{(⌊-log₂(1-t)⌋)} dd
• qed
• Again, desmos calculates this as the integral being 0.916666666667, therefore t_Zeno1 = 24 m * 0.916666666667 = 22 m.

And for extra credit:

Desmos 2

Again, let’s look at what we know:

• v(t) is the speed at any given time.
• v(0) = v(1-1) = v₀ above.
• v(0.5) = v(1-1/2¹) = v₀ * a
• v(0.75) = v(1-1/2²) = v₀ * a²
• v(0.875) = v(1-1/2³) = v₀ * a³
• In general v(1-1/2ⁱ) = v₀ * aⁱ
• And v(t) is a smooth function.
• This calls for a double log system!
• Let’s try simply plugging in some numbers first in Desmos. Looks good.
• Then let’s try to describe this function.
• v(t) = v_opp(1-t) (running the opposite way)
• v_opp(1/2ⁿ) = v₀ * aⁿ
• v_opp(log₂(t)) = v₀ * a^-log₂(t)
• v(t) = v_opp(1-t) = v₀ * a^-log₂(1-t)
• Double log confirms, just writing it as v(t) = v₀ * (1-t)^-0.137504 instead.
• t_Zeno2 = t_init * ∫ ₀¹ 1 / a^-log₂(1-t) dd
• The value of the integral is 0.879118155787, therefore t_Zeno1 = 24 * 0.879118155787 = 21.0988.

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Can You Race Against Zeno?